Trace transfer in a quadratic algebra #
If x generates a degree-two algebra over a field and x² = d, the basis (1, x)
identifies the trace transfer of the unit line with ⟨2, 2d⟩. This supplies the
general degree-two version of the trace-form calculation, with an explicit isometry
whose coordinates can be used to compute invariants of transferred forms.
The trace of x vanishes by Algebra.IsQuadraticExtension.trace_eq_zero_of_sq_eq, which
also covers split and nonreduced algebras.
For a quadratic field extension the file also diagonalizes the twisted trace forms
Tr_*⟨a⟩ : y ↦ Tr (a y²), the transfers of the rank-one forms ⟨a⟩ that Kahn's formula for the
Stiefel–Whitney classes of a transferred form evaluates. There are two cases.
- If
Tr a ≠ 0, Kahn's basis(1, x / a)is orthogonal, andTr_*⟨a⟩ ≅ ⟨Tr a, d · Tr a / N a⟩(equivalent_traceTransfer_smul_sq_weightedSumSquares_of_sq). - If
Tr a = 0anda ≠ 0, then1anda⁻¹are isotropic andTr_*⟨a⟩is the hyperbolic plane (equivalent_traceTransfer_smul_sq_hyperbolicPlane).
References #
- W. Scharlau, Quadratic and Hermitian Forms (1985), Chapter 2, §5.
- B. Kahn, Classes de Stiefel-Whitney de formes quadratiques et de représentations galoisiennes réelles, Invent. Math. 78 (1984), 223–256.
In square-root coordinates, the trace transfer of the unit line is ⟨2, 2d⟩.
The square-root coordinate map is an isometry from the transferred unit line
to the diagonal form ⟨2, 2d⟩.
Instances For
The isometry uses coordinates in the square-root basis.
The inverse isometry reconstructs an element from its square-root coordinates.
The trace transfer of the unit line in a degree-two algebra generated by
x² = d is equivalent to ⟨2, 2d⟩.
Kahn's basis. Let x ∉ K generate a quadratic extension L/K with x² = d, and let
a : L have nonzero trace. Then 1 and x / a are orthogonal for the twisted trace form
Tr_*⟨a⟩ : y ↦ Tr (a y²), with values Tr a and Tr (d / a) = d · Tr a / N a, so
Tr_*⟨a⟩ ≅ ⟨Tr a, d · Tr a / N a⟩.
The hypothesis Tr a ≠ 0 cannot be dropped: at Tr a = 0 both entries vanish, while for
a ≠ 0 the form is the hyperbolic plane (equivalent_traceTransfer_smul_sq_hyperbolicPlane).
The twisted trace form at trace zero. If L/K is a quadratic extension and a ≠ 0 has
trace zero, then Tr_*⟨a⟩ : y ↦ Tr (a y²) is the hyperbolic plane. Indeed 1 and a⁻¹ are
isotropic, as Tr (a⁻¹) = Tr a / N a = 0, and their polar pairing is 2 Tr 1 = 4.